Chapter 4, Problem 7CT

To determine

To graph: The function and label all intercepts, vertical asymptotes, horizontal asymptotes and oblique asymptotes where the function $r\left(x\right)=\frac{x^2+2x-3}{x+1}$ .

Explanation of Solution

Given information:

The function, $r\left(x\right)=\frac{x^2+2x-3}{x+1}$

Graph:

Consider $y=r\left(x\right)=\frac{x^2+2x-3}{x+1}$ .

To find the $x-$ intercept of rational function substitute $y=0$ in the $y=\frac{x^2+2x-3}{x+1}$ ,

  $\Rightarrow 0=\frac{x^2+2x-3}{x+1}$

  $\Rightarrow x^2+2x-3=0$

By using factor method,

  $\Rightarrow \left(x-1\right)\left(x+3\right)=0$

By using zero product rule,

  $\Rightarrow \left(x-1\right)=0$ and $\left(x+3\right)=0$

By simplifying,

  $\Rightarrow x=1$ and $x=-3$ .

Therefore, the $x-$ intercepts are $\left(1,0\right)$ and $\left(-3,0\right)$ .

To find the $y-$ intercept of rational function substitute $x=0$ in the $y=\frac{x^2+2x-3}{x+1}$ ,

  $\Rightarrow y=\frac{{\left(0\right)}^2+2\left(0\right)-3}{\left(0\right)+1}$

  $\Rightarrow y=\frac{-3}{1}=-3$

Therefore, the $y-$ intercept of rational function is $\left(0,-3\right)$ .

To find the asymptotes of the rational function $r\left(x\right)$ ,

Here the rational function $r\left(x\right)=\frac{x^2+2x-3}{x+1}$ is not in the lowest term.

Firstly, convert the rational function in the lowest term.

The numerator $x^2+2x-3$ can factor as $\left(x-1\right)\left(x+3\right)$ and the denominator $x+1$ .

  $\Rightarrow r\left(x\right)=\frac{x^2+2x-3}{x+1}$

  $r\left(x\right)=\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)}$

If $x-a$ is a factor of the denominator of $r\left(x\right)$ , in lowest terms, then $r\left(x\right)$ will have the vertical asymptote $x=a$ .

Here, $x+1$ is a factor of the denominator of $r\left(x\right)$ , then $x=-1$ is a vertical asymptote.

Here the degree of numerator $\left(x-1\right)\left(x+3\right)$ is two and the degree of denominator $x+1$ is one, that means the degree of the numerator is one more than the degree of the denominator therefore the asymptote is oblique asymptote of the form $y=mx+b$ .

The oblique asymptote for the rational function is the quotient of the polynomial division.

To find the oblique asymptote by using the polynomial division,

  $\begin{array}{l}x+1\\ x+1\overline{)x^2+2x-3}\\ \underset{\_}{-x^2+x}\\ x-3\\ \underset{\_}{-x+1}\\ -4\end{array}$

Here the quotient is $x+1$ and remainder is $-4$ .

Therefore, the oblique asymptote for the rational function is $y=x+1$ .

The graph of rational function with $x-$ intercept are $\left(1,0\right)$ and $\left(-3,0\right)$ , the $y-$ intercept $\left(0,-3\right)$ and a vertical asymptote $x=-1$ and the oblique asymptote $y=x+1$ is as shown below;

  

Interpretation:

The graph of rational function with $x-$ intercept are $\left(1,0\right)$ and $\left(-3,0\right)$ , the $y-$ intercept $\left(0,-3\right)$ and a vertical asymptote $x=-1$ and the oblique asymptote is $y=x+1$ .